University Physics I - Classical Mechanics, 2019

8.7. ADVANCED TOPICS 179
8.7 Advanced Topics
8.7.1 Staying on track
(This example studies a situation that you could easily setup experimentally at home (you can use
a whole sphere instead of a half-sphere!), although to get the numbers to work out you really need
to make sure that the friction between the surface and the object you choose is truly negligible.
Essentially the same mathematical approach could be used to study the problem of a skier going
over a mogul, or a car losing contact with the road if it is going too fast over a hill.)
A small object is placed at the top of a smooth (frictionless) dome shaped like a half-sphere of
radius R, and given a small push so it starts sliding down the dome, initially moving very slowly
(v i ≃ 0), but picking up speed as it goes, until at some point it flies off the surface.
(a) At that point, when the object loses contact with the surface, what is the angle that its position
vector (with origin at the center of the sphere) makes with the vertical?
(b) How far away from the sphere does the object land?
F n
sb
θ
θ max
θ
θ
F G
Eb
(a)
(b)
Figure 8.8: An object (small block) sliding on a hemispherical dome. The drawing (a) shows the angle
θ max at which the object flies off (red dashed line), and a smaller, generic angle θ. The drawing (b) shows
the free-body diagram corresponding to the angle θ.
Solution
(a) As we saw in Section 8.4, in order to get an object to move along an arc of a circle, a centripetal
force of magnitude mv 2 /r is required. As long as our object is in contact with the surface, the
forces acting on it are the normal force (which points along the radial direction, so it makes a
negative contribution to the centrifugal force) and gravity, which has a component mg cos θ along
the radius, towards the center of the circle (see Figure 8.8(b), the dashed light blue line). So, the
centripetal force equation reads
mv 2
R = mg cos θ − F n (8.46)